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Proof by mathematical induction

Let u1 = 1,...

Question

Let $u_{1}$ = 1, $u_{2}$ = 1 and $u_{n + 2} + u_{n} f o r n \geq 1$ .
Use mathematical induction to show that
$u_{n} = \frac{1}{\sqrt{(5)}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}]$ for all $n \geq 1$

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Solution

We have to prove
$u_{n} = \frac{1}{\sqrt{(5)}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}]$
for all n $\geq$ 1
We obviously have
$u_{1} = 1 = \frac{1}{\sqrt{(5)}} [\frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2}]$
and $u_{2} = 1 = \frac{1}{\sqrt{(5)}} ⎡ ⎣ {(\frac{1 + \sqrt{5}}{2})}^{2} - {(\frac{1 - \sqrt{5}}{2})}^{2} ⎤ ⎦$
Hence (1) holds for n = 1 and n = 2
Now assume
$u_{k} = \frac{1}{\sqrt{(5)}} ⎡ ⎣ {(\frac{1 + \sqrt{5}}{2})}^{k} - {(\frac{1 - \sqrt{5}}{2})}^{k} ⎤ ⎦$
( k = 1, 2, 3, ....m)
Now $u_{m + 2} = u_{m + 1} + u_{m}$ for m $\geq$ 1
$\Rightarrow u_{m + 1} = u_{m} + u_{m - 1}$ for m $\geq$ 2
Hence by induction hypothesis on $u_{k}$ , we have
$u_{m + 1} = u_{m} + u_{m - 1}$
$= \frac{1}{\sqrt{(5)}} [{(\frac{1 + \sqrt{5}}{2})}^{m} - {(\frac{1 - \sqrt{5}}{2})}^{m}] + \frac{1}{\sqrt{(5)}} ⎡ ⎣ {(\frac{1 + \sqrt{5}}{2})}^{m - 1} - {(\frac{1 - \sqrt{5}}{2})}^{m - 1} ⎤ ⎦$
$= \frac{1}{\sqrt{(5)}} ⎡ ⎣ {(\frac{1 + \sqrt{5}}{2})}^{m - 1} {\frac{1 + \sqrt{5}}{2} + 1} - {(\frac{1 - \sqrt{5}}{2})}^{m - 1} {\frac{1 - \sqrt{5}}{2} + 1} ⎤ ⎦$
$= \frac{1}{\sqrt{(5)}} ⎡ ⎣ {(\frac{1 + \sqrt{5}}{2})}^{m - 1} (\frac{6 + 2 \sqrt{5}}{4}) ⎤ ⎦ - {(\frac{1 - \sqrt{5}}{2})}^{m - 1} (\frac{6 - 2 \sqrt{5}}{4})$
$= \frac{1}{\sqrt{(5)}} ⎡ ⎣ {(\frac{1 + \sqrt{5}}{2})}^{m - 1} (\frac{1 + \sqrt{5}}{2}) ⎤ ⎦ - ⎡ ⎣ {(\frac{1 - \sqrt{5}}{2})}^{m - 1} (\frac{1 - \sqrt{5}}{2}) ⎤ ⎦$
$= \frac{1}{\sqrt{(5)}} ⎡ ⎣ {(\frac{1 + \sqrt{5}}{2})}^{m + 1} {(\frac{1 - \sqrt{5}}{2})}^{m + 1} ⎤ ⎦$
Thus the formula (1) hold for k = m + 1
Hence(1)holds for all positive integers n by induction

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